# A regular triangle is inscribed in a circle. Find the area of the smaller segment of the cut

**A regular triangle is inscribed in a circle. Find the area of the smaller segment of the cut off one of the sides of the triangle if the radius of the circle is 18 cm.**

Determine the area of the circle. Scr = π * R2 = π * 324 cm2.

The radii ОА, ОВ, OC divide the circle into three equal sectors, then the area of the AOB sector is equal to:

Ssec = Scr / 3 = π * 324/3 = π * 108 cm2.

In triangle AOB, angle AOB = 360/3 = 120. Triangle AOB is isosceles, since OA = OB = 18 cm, then the height OH is also its median and bisector, then angle BOH = 120/2 = 60, angle OBH = (90 – 60) = 30. Then the leg OH lies opposite the angle 30, then OH = OB / 2 = 18/2 = 9 cm.

BH ^ 2 = OB ^ 2 – OH ^ 2 = 324 – 81 = 243.

BH = 9 * √3 cm.Then AB = 2 * BH = 18 * √3 cm.

Determine the area of the triangle ABC.

Savs = AB * OH / 2 = 18 * √3 * 9/2 = 81 * √3 cm2.

The area of the segment is equal to: Sseg = Ssec – Savs = π * 108 – 81 * √3 = 27 * (4 * π – 3 * √3) cm2.

Answer: The area of the segment is 27 * (4 * π – 3 * √3) cm2.